Hilbert's Type Linear Operator and Some Extensions of Hilbert's Inequality

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Volume 2007, Article ID 82138,10pages doi:10.1155/2007/82138

Research Article

Hilbert’s Type Linear Operator and Some Extensions of

Hilbert’s Inequality

Yongjin Li, Zhiping Wang, and Bing He

Received 17 April 2007; Accepted 3 October 2007

Recommended by Ram N. Mohapatra

The norm of a Hilbert’s type linear operatorT:L2_(0,_∞₎_→_L2_(0,_∞_{) is given. As}

appli-cations, a new generalizations of Hilbert integral inequality, and the result of series ana-logues are given correspondingly.

Copyright © 2007 Yongjin Li et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1. Introduction

At the close of the 19th century a theorem of great elegance and simplicity was discovered by D. Hilbert.

Theorem1.1 (Hilbert’s double series theorem). The series

∞

m=1

∞

n=1

aman

m+n (1.1)

is convergent whenever∞_n₌1a2nis convergent.

The Hilbert’s inequalities were studied extensively; refinements, generalizations, and numerous variants appeared in the literature (see [1, 2]). Firstly, we will recall some Hilbert’s inequalities. If f(x),g(x)≥0, 0<0∞f2(x)dx <∞and 0<

_∞

0 g2(x)dx <∞, then

_∞

0

_∞

0

f(x)g(y)

x+y dx d y < π

_∞

0 f 2_(x)dx

1/2_∞

0 g 2_(x)dx

1/2

, (1.2)

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Hardy-Hilbert’s type inequality (see [3], Theorem 319, Theorem 341) similar to (1.2):

_∞

0

_∞

0

f(x)g(y)

max{x,y}dx d y <4

_∞

0 f 2_(x)dx

1/2_∞

0 g 2_(x)dx

1/2

, (1.3)

where the constant factor 4 is also the best possible. The corresponding inequalities for series are:

∞

n=1

∞

m=1

ambn

m+n< π _∞

n=1

a2

n

1/2_∞

n=1

b2

n

1/2

;

∞

n=1

∞

m=1

ambn

max{m,n}<4

_∞

n=1

a2

n

1/2_∞

n=1

b2

n

1/2

,

(1.4)

where the constant factorsπand 4 are both the best possible.

LetHbe a real separable Hilbert space, andT:H→Hbe a bounded self-adjoint semi-positive definite operator, then (see [4])

(x,T y)2≤T 2

2

x2_y2_{+ (x,}_y)2

, (1.5)

wherex,y∈Handx =(x,x) is the norm ofx. SetH=L2_(0,_∞₎_{= {}_f_{(x) :}∞

0 f2(x)dx <∞}and defineT:L2(0,∞)→L2(0,∞) as the

following:

(T f)(y) :=

_∞

0

1

x+yf(x)dx, (1.6)

wherey∈(0,∞). It is easy to seeTis a bounded operator (see [5]). By (1.5), one has the sharper form of Hilbert’s inequality as (see [4]),

_∞

0

_∞

0

f(x)g(y)

x+y dx d y≤ π

√

2

_∞

0 f 2_(x)dx

_∞

0 g

2_(x)dx₊

_∞

0 f(x)g(x)dx 21/2

.

(1.7)

Recently, Yang [6,7] studied the Hilbert’s inequalities by the norm of some Hilbert’s type linear operators.

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2. Main results and applications

Lemma2.1. Define the weight function(x)as

(x) :=

∞

0

1

Amin{x,y}+Bmax{x,y}

x y

1/2

d y, x∈(0,∞),

(y)

_∞

0

1

Amin{x,y}+Bmax{x,y}

y x

1/2

dx, y∈(0,∞).

(2.1)

Then(x)=(y)=D(A,B)is a constant and0< D(A,B)<∞. In particular, one hasD(1, 1)=πandD(1, 0)=4.

Proof. For fixedx, lettingt=y/x, we get

(x)=

_∞

0

1

Amin{x,y}+Bmax{x,y}

x y

1/2

d y

=

_∞

0

1

Amin{1,t}+Bmax{1,t}t−1/2dt

=

1

0

1 At+Bt

−1/2_dt₊

_∞

1

1 A+Btt

−1/2_dt

=√1

AB _A/B

0

1 1 +tt

−1/2_dt₊_√1

AB _∞

B/A

1 1 +tt

−1/2_dt

≤√1

AB _∞

0

1 1 +tt

−1/2_dt₊_√1

AB _∞

0

1 1 +tt

−1/2_dt

=√2

ABB

1 2,

1 2 <∞.

(2.2)

therefore 0< D(A,B)<∞. Moreover,

(y)=

_∞

0

1

Amin{x,y}+Bmax{x,y}

y x

1/2

dx

=

_∞

0

1

Amin{1,t}+Bmax{1,t}t−1/2dt

=

1

0

1 At+Bt−

1/2_dt₊

_∞

1

1 A+Btt−

1/2_dt

= √1

AB

A−1+(1/2)

B1/2

A/B

0

1 1 +tt

−1/2_dt₊_√1

AB ∞

B/A

1 1 +tt

−1/2_dt

=√1

AB A/B

0

1 1 +uu−

1/2_du₊_√1

AB _∞

B/A

1 1 +uu−

1/2_du

(2.3)

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Thus(y)=D(A,B). In particular:

D(1, 1)=

∞

0

1 x+y

y x

1/2

dx=

∞

0

1 1 +tt

−1/2_dt₌_π,

D(0, 1)=

_∞

0

1 max{x,y}

y x

1/2

dx=

_∞

0

1 max{1,t}t−

1/2_dt₌_4.

(2.4)

Theorem2.2. LetA≥0,B >0andT:L2_(0,_∞₎_→_L2_(0,_∞₎_{is defined as follows:}

(T f)(y) :=

_∞

0

1

Amin{x,y}+Bmax{x,y}f(x)dx (y∈(0,∞)). (2.5)

Then T =D(A,B), and for any f(x),g(x)≥0, f,g ∈L2_(0,_∞₎_{, one has} _{(T f}_,g₎_<

D(A,B)fg, that is,

_∞

0

_∞

0

f(x)g(y)

Amin{x,y}+Bmax{x,y}dx d y < D(A,B)

_∞

0 f 2_(x)dx

1/2_∞

0 g 2_(x)dx

1/2

,

(2.6)

where the constant factorD(A,B)is the best possible. In particular, (i)forA=B=1, it reduces to Hardy-Hilbert’s inequality:

_∞

0

_∞

0

f(x)g(y)

x+y dx d y < π

_∞

0 f 2_(x)dx

1/2_∞

0 g 2_(x)dx

1/2

; (2.7a)

(ii)forA=0,B=1, it reduces to Hardy-Hilbert’s type inequality:

_∞

0

_∞

0

f(x)g(y)

max{x,y}dx d y <4

_∞

0 f 2_(x)dx

1/2_∞

0 g 2_(x)dx

1/2

. (2.7b)

Proof. ForA >0,B >0. Applying H¨older’s inequality, we obtain

(T f,g)=

_∞

0

f(x)

Amin{x,y}+Bmax{x,y}dx,g(y)

=

_∞

0

_∞

0

f(x)

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=

_∞

0

_∞

0

f(x)g(y)

Amin{x,y}+Bmax{x,y}dx d y

=

_∞

0

_∞

0

1

Amin{x,y}+Bmax{x,y}

f(x)

x y

1/4

g(y)

y x

1/4

dx d y

≤

_∞

0

_∞

0

f2_(x)

Amin{x,y}+Bmax{x,y}

x y

1/2

dx d y 1/2

×

_∞

0

∞

0

g2_(y)

Amin{x,y}+Bmax{x,y}

y x

1/2

dx d y 1/2

=

_∞

0 (x)f 2_(x)dx

1/2_∞

0 (y)g 2_{(y)d y}

1/2

=D(A,B)

_∞

0 f 2_(x)dx

1/2_∞

0 g 2_{(y)d y}

1/2

=D(A,B)fg.

(2.8)

ThusT ≤D(A,B) and the inequality (2.6) holds.

Assume that (2.8) takes the form of the equality, then there exist constantsaandb, such that they are not both zero and (see [8])

a f2(x)

x y

1/2

=bg2(y)

y x

1/2

. (2.9)

Then, we have

a f2_(x)x₌_bg2_(y)y _a.e. _{on (0,}_∞₎_×_(0,_∞_). _(2.10)

Hence there exist a constantd, such that

a f2(x)x=bg2(y)y =d a.e.on (0,∞)×(0,∞). (2.11)

Without losing the generality, suppose a =0, then we obtain f2_(x)₌_{d/(ax), a.e. on}

(0,∞), which contradicts the fact that 0<0∞f2(x)dx <∞. Hence (2.8) takes the form

of strict inequality, we obtain (2.6).

Forε >0 suﬃciently small, set fε(x)=x(−1−ε)/2, forx∈[1,∞);fε(x)=0, forx∈(0, 1).

Thengε(y)=y(−1−ε)/2, fory∈[1,∞);gε(y)=0, fory∈(0, 1). Assume that the constant

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withK < D(A,B), such that (2.6) is valid by changingD(A,B) toK. On one hand,

_∞

0

_∞

0

f(x)g(y)

Amin{x,y}+Bmax{x,y}dx d y < K

_∞

0 f 2

ε(x)dx

1/2_∞

0 g 2

ε(x)dx

1/2 =K/ε.

(2.12)

On the other hand, settingt=y/x, we have _∞

0

_∞

0

fε(x)gε(y)

=

_∞

1

_∞

1

x(−1−ε)/2_y(−1−ε)/2

=

_∞

1 x

−1−ε

_∞

1/x

t(−1−ε)/2

Amin{1,t}+Bmax{1,t}dt dx

=

∞

1 x

−1−ε

∞

0

t(−1−ε)/2

Amin{1,t}+Bmax{1,t}dt dx

−

_∞

1 x

−1−ε

1/x

0

t(−1−ε)/2

Amin{1,t}+Bmax{1,t}dt dx.

(2.13)

Forx≥1, we get

1/x

0

t(−1−ε)/2

Amin{1,t}+Bmax{1,t}dt

=

1/x

0

t(−1−ε)/2

At+B dt

≤ 1

B 1/x

0 t

(−1−ε)/2_dt

=_B1 1

1−(1 +ε)/2

1 x

1−(1+ε)/2

≤ 4

Bx

−1/4

(2.14)

(setting 0< ε <1/2). Thus

0< ∞

1 x

−1−ε1/x

0

t(−1−ε)/2

Amin{1,t}+Bmax{1,t}dt dx

≤4

B ∞

1 x

−1−ε−1/4_dx

≤4

B _∞

1 x

−1−1/4_dx₌16

B.

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Note that

_∞

1 x

−1−ε

1/x

0

t(−1−ε)/2

Amin{1,t}+Bmax{1,t}dt dx=O(1). (2.16)

So the inequality0∞

∞

0 (fε(x)gε(y)/(Amin{x,y}+Bmax{x,y}))dxd y=(1/ε)[D(A,B) +

o(1)]−O(1)=(1/ε)[D(A,B) +o(1)]. Thus we get (1/ε)[D(A,B,p) +o(1)]≤K/ε, that is, D(A,B)≤K when ε is suﬃciently small, which contradicts the hypothesis. Hence the constant factorD(A,B) in (2.6) is the best possible andT =D(A,B). This completes

the proof.

Theorem2.3. Suppose that f ≥0,A≥0,B >0and 0<0∞f2(x)dx <∞. Then

_∞

0

_∞

0

f(x)

Amin{x,y}+Bmax{x,y}dx

2

d y < D2_(A,B)

_∞

0 f

2_(x)dx, _(2.17)

where the constant factorD2_(A,B)_{is the best possible. Inequality (2.17) is equivalent to (2.6).}

Proof. Letg(y)=0∞(f(x)/(Amin{x,y}+Bmax{x,y}))dx, then by (2.6), we get

0< _∞

0 g 2_{(y)d y}

=

_∞

0

_∞

0

f(x)

Amin{x,y}+Bmax{x,y}dx

2

d y

=

_∞

0

_∞

0

f(x)g(y)

≤D(A,B)

_∞

0 f 2_(x)dx

1/2_∞

0 g 2_{(y)d y}

1/2

.

(2.18)

Hence, we obtain

0< _∞

0 g

2_{(y)d y}₌_D2_(A,_B)

_∞

0 f

2_{(x)dx <}_∞_. _(2.19)

By (2.6), both (2.18) and (2.19) take the form of strict inequality, so we have (2.17). On the other hand, suppose that (2.17) is valid. By H¨older’s inequality, we find

_∞

0

_∞

0

f(x)g(y)

=

_∞

0

_∞

0

f(x)

Amin{x,y}+Bmax{x,y}dx

g(y)d y

≤

_∞

0

_∞

0

f(x)

Amin{x,y}+Bmax{x,y}dx

2

d y 1/2

∞

0 g 2_(x)dx

1/2

.

(2.20)

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If the constant D2_{(A,B) in (}_2.17_{) is not the best possible, by (}_2.20_{), we may get a}

contradiction that the constant factor in (2.6) is not the best possible. This completes the

proof.

It is easy to see that forA=1,B=1, the inequality (2.17) reduces to

_∞

0

_∞

0

f(x) x+ydx

2

d y < π2

_∞

0 f

2_(x)dx, _(2.21a)

and forA=0,B=1, the inequality (2.17) reduces to

_∞

0

_∞

0

f(x) max{x,y}dx

2

d y <16 _∞

0 f

2_(x)dx, _(2.21b)

where both the constant factorsπ2_{and 16 are the best possible.}

3. The corresponding theorem for series

Theorem3.1. Suppose that an,bn≥0,A≥0,B >0, and 0<∞n=1a2n<∞,0<

_∞

n=1b2n<

∞. Then

∞

n=1

∞

m=1

ambn

Amin{m,n}+Bmax{m,n}< D(A,B)

_∞

n=1

a2

n

1/2_∞

n=1

b2

n

1/2

, (3.1)

∞

n=1

_∞

m=1

am

Amin{m,n}+Bmax{m,n}

2

< D2(A,B)

∞

n=1

a2_n, (3.2)

where the constant factorD(A,B)andD2_(A,B)_{are both the best possible, (3.1) and (3.2)}

are equivalent. In particular,

(i)forA=1,B=1, it reduces to Hardy-Hilbert’s inequality:

∞

n=1

∞

m=1

ambn

m+n< π _∞

n=1

a2

n

1/2_∞

n=1

b2

n

1/2

; (3.3a)

(ii)forA=0,B=1, it reduces to Hardy-Hilbert’s type inequality:

∞

n=1

∞

m=1

ambn

max{m,n}<4

_∞

n=1

a2n

1/2_∞

n=1

b2n

1/2

. (3.3b)

Proof. Define the weight functionω(n) as

ω(n) :=

∞

m=1

1

Amin{m,n}+Bmax{m,n}

n m

1/2

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Then we obtain

ω(n)< (n)=D(A,B). (3.5)

Using the method similar toTheorem 2.2and applying H¨older’s inequality, we obtain

∞

n=1

∞

m=1

ambn

Amin{m,n}+Bmax{m,n}≤

_∞

n=1

ω(n)a2

n

1/2_∞

n=1

ω(n)b2

n

1/2

. (3.6)

By (3.5), we obtain (3.1).

Forε >0 suﬃciently small, settingan=n−(1+ε)/2,bn=n−(1+ε)/2, then

∞

n=1

∞

m=1

ambn

Amin{m,n}+Bmax{m,n}>

_∞

1

_∞

1

fε(x)gε(y)

Amin{x,y}+Bmax{x,y}dx d y,

_∞

n=1

a2

n

1/2_∞

n=1

b2

n

1/2 =

∞

n=1

1 n1+ε<1 +

∞

1

1 t1+ε =1 +

1 ε.

(3.7)

If the constant factor D(A,B) in (3.1) is not the best possible, then applying the re-sult ofTheorem 2.2, we can get the contradiction. Letbn=∞m=1(am/(Amin{m,n}+

Bmax{m,n})) and we can obtain the following relation:

∞

n=1

_∞

m=1

am

Amin{m,n}+Bmax{m,n}

2

=

∞

n=1

b2n= ∞

n=1

∞

m=1

ambn

Amin{m,n}+Bmax{m,n}.

(3.8)

Applying (3.1) and the method similar toTheorem 2.3, we get (3.2), and (3.2) is equiva-lent to (3.1) with the best constant.

Acknowledgments

The work was partially supported by the Emphases Natural Science Foundation of Guang-dong Institution of Higher Learning, College and University (no. 05Z026). The authors would like to thank the anonymous referee for his or her suggestions and corrections.

References

[1] E. F. Beckenbach and R. Bellman,Inequalities, Springer, Berlin, Germany, 1961. [2] D. S. Mitrinovi´c,Analytic Inequalities, Springer, New York, NY, USA, 1970.

[3] G. H. Hardy, J. E. Littlewood, and G. P ´olya, Inequalities, Cambridge University Press, Cam-bridge, UK, 2nd edition, 1952.

[4] K. Zhang, “A bilinear inequality,”Journal of Mathematical Analysis and Applications, vol. 271, no. 1, pp. 288–296, 2002.

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[6] B. Yang, “On the norm of an integral operator and applications,”Journal of Mathematical Anal-ysis and Applications, vol. 321, no. 1, pp. 182–192, 2006.

[7] B. Yang, “On the norm of a Hilbert’s type linear operator and applications,”Journal of Mathe-matical Analysis and Applications, vol. 325, no. 1, pp. 529–541, 2007.

[8] J. C. Kuang,Changyong Budengshi, Hunan Jiaoyu Chubanshe, Changsha, China, 2nd edition, 1993.

Yongjin Li: Institute of Logic and Cognition, Department of Mathematics, Sun Yat-Sen University, Guangzhou 510275, China

Email address:[email protected]

Zhiping Wang: Department of Mathematics, Guangdong University of Finance, Zhaoqing Campus, Zhaoqing 526060, China

Email address:[email protected]

Bing He: Department of Mathematics, Guangdong Education College, Guangzhou 510303, China